First, evaluate the integral:
color(white)=int((3x+1)^2-2x-1) dx
=int(3x+1)^2 dx -int2x dx -int1 dx
=int(3x+1)^2 dx -(2x^2)/2-x+C
=int(3x+1)^2 dx -x^2-x+C
Let u=3x+1, which means du=3dx, or dx=(du)/3:
=intu^2/3 du -x^2-x+C
=1/3intu^2 du -x^2-x+C
=1/3*u^3/3-x^2-x+C
=u^3/9-x^2-x+C
=(3x+1)^3/9-x^2-x+C
=(27x^3+27x^2+9x+1)/9-x^2-x+C
=3x^3+3x^2+x+1/9-x^2-x+C
=3x^3+2x^2+C
=x^2(3x+2)+C
Now, plug in x=2 and set this equal to 5 and solve for C:
2^2(3(2)+2)+C=5
4(8)+C=5
32+C=5
C=-27
This is our set constant. This means that our f(x) is:
f(x)=x^2(3x+2)-27
Hope this helped!